What is lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1)) ?

3 Answers
Mar 25, 2016

The limit is -1/sqrt2

Explanation:

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We need to find the asymptote to sqrt(n^2+0.5n+1).
(Thanks, George C.)

To find (linear) oblique asymptote y=mn+b,

m = lim_(nrarroo) sqrt(n^2+0.5n+1)/n = 1.

Now we find b:

b= lim_(nrarroo) sqrt(n^2+0.5n+1)/mn = lim_(nrarroo) sqrt(n^2+0.5n+1)/n = 1/4.

So, sqrt(n^2+0.5n+1) is asymptotic to n+1/4.

Alternative method from George C

sqrt(n^2+0.5n+1) = sqrt((n+1/4)^2+15/16)

= (n+1/4)sqrt(1+15/(16(n+1/4)^2)

(If I've deduced the method correctly, our goal is to get (mn+b)sqrt(1+g(n)) where lim_(nrarroo)g(n)=0.)

Returning to the big question

lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1)) = lim_(nrarroo)(-1)^(n-1)sin(pi(n+1/4)sqrt(1+15/(16(n+1/4)^2)))

For even n, say n=2k,
we get -sin(2kpi+pi/4) = -sin(pi/4) = -1/sqrt2

For odd n, say n=2k+1,
we get sin(2kpi+pi+pi/4) = sin((5pi)/4) = -1/sqrt2

Therefore,

lim_(nrarroo)(-1)^(n-1)sin(pisqrt(n^2+0.5n+1)) = -1/sqrt2

Mar 25, 2016

-sqrt2/2

Explanation:

Extra care should be taken since we are dealing with a periodic function of the kind
f(x)=+-sinx

It's not sufficient that, when n->oo, n^2 is much greater than 0.5n+1, because if sqrt(n^2+0.5n+1)-n > 0 than the function will have a value possibly different than zero.

Intuitively it's easy to see that sqrt(k+1)->sqrt k (or, even more importantly for the case, that sqrt(k+1)-sqrt k->0) when k->oo
Or
sqrt(1001)-sqrt(1000)=0.015,807
and
sqrt(1,000,001)-sqrt(1,000,000)=0.000,500
and so on.

But let's prove that sqrt(n^2+0.5n+a)->sqrt(n^2+0.5n) when n->oo, where n in NN and a in RR
lim_(n->oo) (sqrt(n^2+0.5n+a)-sqrt(n^2+0.5n))*((sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n)))/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))=
=lim_(n->oo) (cancel(n^2)+cancel(0.5n)+a-cancel(n^2)-cancel(0.5n))/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))
=lim_(n->oo) a/(sqrt(n^2+0.5n+a)+sqrt(n^2+0.5n))
=lim_(n->oo) a/oo=0

Since the value of a (in the present case a=1) doesn't matter to the result, we can make
lim_(n->oo) sqrt(n^2+0.5n+1)=lim_(n->oo) sqrt(n^2+0.5n+1/16)=lim_(n->oo) sqrt((n+1/4)^2)=lim_(n->oo) (n+1/4)

So the main expression becomes

=lim_(n->oo) (-1)^(n-1)sin[pi(n+1/4)]
=lim_(n->oo) (-1)^(n-1)sin(n*pi+pi/4)
But
sin(n*pi+pi/4)=sin(n*pi)*cos(pi/4)+sin(pi/4)*cos(n.pi)=
(remembering that n in NN => sin(n*pi)=0)
=sqrt2/2*cos(n*pi)
In the main expression
=lim_(n->oo) (-1)^(n-1)*sqrt2/2*cos(n*pi)

Now consider the possible values of the expression above
If n is odd => (-1)^("even number").sqrt2/2*cos[("odd number")*pi]=1*sqrt2/2(-1)=-sqrt2/2
If n is even => (-1)^("odd number").sqrt2/2*cos[("even number")*pi]=(-1)*sqrt2/2*1=-sqrt2/2

So, for n in NN
=lim_(n->oo) (-1)^(n-1)*sqrt2/2*cos(n*pi)=-sqrt2/2

Mar 27, 2016

-1/sqrt2.

Explanation:

I have revised my answer, thanks to George for his affirmation on his answer. I am sorry for having overlooked some nicety in applying the notion of 'one is asymptotic with another', in this limit problem.

The correction follows, without changing my approach.

pisqrt(n^2+0.5n+1)=npisqrt(1+0.5/n+1/n^2))=npi(1+(1/2)(0.5/n)+O(1/n^2)), using expansion in powers of 1/n.
Here, ntooo through integer values 1,2,3,..,

The limit is same as the limit for
(-1)^nsin(npi+pi/4+O(1/n)) =-1/sqrt2, as given in the other answers..